B=\(\dfrac{1+15^4+15^8+...+15^{96}+15^{100}}{\left(1+15^4+15^8+..+15^{96}+15^{100}\right)+\left(15^2+15^6+...+15^{98}+15^{102}\right)}\)

=\(\dfrac{1+15^4+15^8+...+15^{96}+15^{100}}{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)+15^2.\left(1+15^{14}+15^8+...+15^{96}+15^{100}\right)}\)

\(\dfrac{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)}{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)\left(1+15^2\right)}\)

=\(\dfrac{1}{1+15^2}=\dfrac{1}{226}\)

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Ủng hộ em vài nha

\(\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}=\frac{15\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}{18\left(1+\frac{1}{7}+\frac{1}{12}+\frac{1}{98}\right)}=\frac{15}{18}=\frac{5}{6}\)

a) \(A=\frac{15^{16}+1}{15^{17}+1}\)và\(B=\frac{15^{15}+1}{15^{16}+1}\)

ta có \(A=\frac{15^{16}}{15^{17}}\)và\(B=\frac{15^{15}}{15^{16}}\)

ta dễ nhận thấy phần cơ số của hai phân số A và B = nhau

mà phần mũ của các lũy thừa phân số A đều lớn hơn phân số B 

\(\Rightarrow\frac{15^{16}}{15^{17}}>\frac{15^{15}}{15^{16}}\)

\(\Rightarrow\frac{15^{16}+1}{15^{17}+1}>\frac{15^{15}+1}{15^{16}+1}\)

\(\Rightarrow A>B\)

\(A=\frac{15^{16}+1}{15^{17}+1}vaB=\frac{15^{15}+1}{15^{16}+1}\)

+)Ta thấy\(A=\frac{15^{16}+1}{15^{17}+1}< 1\)

\(\Rightarrow A< \frac{15^{16}+1+14}{15^{17}+1+14}=\frac{15^{16}+15}{15^{17}+15}=\frac{15.\left(15^{15}+1\right)}{15.\left(15^{15}+1\right)}=\frac{15^{15}+1}{15^{16}+1}=B\)

Vậy A<B

b)Đề sai

Chúc bạn học tốt

Ta có: \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)

= \(15.\left(\frac{1}{90.94}+\frac{1}{94.98}+\frac{1}{98.102}+...+\frac{1}{146+150}\right)\)

= \(15.\left[\frac{1}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+\frac{4}{98.102}+...+\frac{4}{146+150}\right)\right]\)

= \(15.\left[\frac{1}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\right]\)

= \(15.\left[\frac{1}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\right]\)

= \(15.\left(\frac{1}{4}.\frac{1}{225}\right)\)

= \(=\frac{1}{60}\)

Bài làm

   \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)

= \(15.\frac{1}{90.94}+15.\frac{1}{94.98}+15.\frac{1}{98.102}+...+15.\frac{1}{146.150}\)

= \(15.\left(\frac{1}{90.94}+\frac{1}{94.98}+\frac{1}{98.102}+...+\frac{1}{146.150}\right)\)

= \(15.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)

= \(15.\left(\frac{1}{90}-\frac{1}{150}\right)\)

= \(15.\left(\frac{5}{450}-\frac{3}{450}\right)\)

= \(15.\frac{2}{450}\)

= \(\frac{2}{30}\)

# Chúc bạn học tốt #

\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}\)

\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)+\left(1-\frac{1}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)

\(=6-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-1+\frac{1}{13}\)

\(=5+\frac{1}{13}\)

\(=\frac{66}{13}\)

Mk sửa lại 1 tí nha dòng thứ 5 :

\(A=6-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}\left(1-\frac{1}{13}\right)\)

\(=6-\frac{1}{2}.\frac{12}{13}\)

\(=6-\frac{6}{13}=\frac{72}{13}\)

Mong bn bỏ qua nha

Ta có \(\frac{-7}{6}=\frac{a-2}{18}\Rightarrow a-2=-21\Rightarrow a=-19\)

\(\frac{-7}{6}=\frac{-98}{3b+15}\Rightarrow3b+15=84\Rightarrow b=23\)

\(\frac{-7}{6}=\frac{14}{-5c+37}\Rightarrow-5c+37=-12\Rightarrow c=\frac{49}{5}\)

\(\frac{-7}{6}=\frac{d+1}{12}\Rightarrow d+1=-14\Rightarrow d=-15\)

bạn xem lai đề phần a đi mik giúp cho